acids and bases
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18-1
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Acids and Bases
18-2
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18-3
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Arrhenius Acid-Base Definition
An acid is a substance that increase H+ when dissolved in water.
A base is a substance that increase OH- ions when dissolved in water.
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
The ionization of acids actually produces H3O+ but H3O+ H2O + H+
NaOH(aq) + H2O(l) Na+(aq) + OH-(aq)
Limitation: this theory cannot explain why NH3 is a base
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Brønsted-Lowry Acid-Base Definition
An acid reactant will produce a base product and the two will constitute an acid-base conjugate pair.
An acid is a proton donor, any species which donates a H+.
A base is a proton acceptor, any species which accepts a H+.
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
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An acid and base that differ only in the presence or absence of a proton
The Conjugate Acid-Base Pairs
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Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions
Base Acid+Acid Base+
Conjugate Pair
Conjugate Pair
Reaction 4 H2PO4- OH-+
Reaction 5 H2SO4 N2H5++
Reaction 6 HPO42- SO3
2-+
Reaction 1 HF H2O+ F- H3O++
Reaction 3 NH4+ CO3
2-+
Reaction 2 HCOOH CN-+ HCOO- HCN+
NH3 HCO3-+
HPO42- H2O+
HSO4- N2H6
2++
PO43- HSO3
-+
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SAMPLE PROBLEM 18.4: Identifying Conjugate Acid-Base Pairs
PROBLEM: The following reactions are important environmental processes. Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO3
2-(aq) HPO42-(aq) + HCO3
-(aq)
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3
-(aq)
SOLUTION:
PLAN: Identify proton donors (acids) and proton acceptors (bases).
(a) H2PO4-(aq) + CO3
2-(aq) HPO42-(aq) + HCO3
-(aq)
proton donor
proton acceptor
proton acceptor
proton donor
conjugate pair1conjugate pair2
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3
-(aq)
conjugate pair2conjugate pair1
proton donor
proton acceptor
proton acceptor
proton donor
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A substance that can both accept and donate a proton– i.e. act as an acid and as a base
Ex. H2O
Amphoteric substances
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
Water acts as a base
NH3 + H2O(l) NH4+(aq) + OH-(aq)
Water acts as an acid
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Figure 18.9
Strengths of conjugate acid-
base pairs
18-10
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In every acid-base reaction, the position of equilibrium favors the weaker acid
Relative Strengths of Acids and Bases
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
Since H3O+ is weaker, the forward reaction is favored over the reverse reaction and the equilibrium lies to the right
Stronger acid Weaker acid
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SAMPLE PROBLEM 18.5: Predicting the Net Direction of an Acid-Base Reaction
PROBLEM: Predict the net direction of equilibrium
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
(a) H2PO4-(aq) + NH3(aq) HPO4
2-(aq) + NH4+(aq)
SOLUTION:
PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10 (button) to determine the relative strength of each. The stronger the species, the more preponderant its conjugate.
(a) H2PO4-(aq) + NH3(aq) HPO4
2-(aq) + NH4+(aq)
stronger acid weaker acid
Net direction is to the right with Kc > 1.
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
stronger acidweaker acidNet direction is to the left with Kc < 1.
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Kc = [H3O+][OH-]
[H3O+][OH-]
The Ion-Product Constant for Water
Kw =
A change in [H3O+] causes an inverse change in [OH-].
= 1.0 x 10-14 at 250C
H2O(l) H3O+(aq) + OH-(aq)
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]
Autoionization of Water
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Figure 18.4The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
[H3O+] [OH-]Divide into Kw
ACIDIC SOLUTION
BASIC SOLUTION
[H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-]
NEUTRAL SOLUTION
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SAMPLE PROBLEM 18.2: Calculating [H3O+] and [OH-] in an Aqueous
Solution
PROBLEM: A research chemist adds a measured amount of HCl gas to pure water at 250C and obtains a solution with [H3O+] = 3.0x10-4M. Calculate [OH-]. Is the solution neutral, acidic, or basic?
SOLUTION:
PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].
Kw = 1.0x10-14 = [H3O+] [OH-] so
[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 =
[H3O+] is > [OH-] and the solution is acidic.
3.3x10-11M
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Solve the following:
a. [ H+] = 1.4 x 10-6 [OH-] =
b. [ OH-] = 1.0 x 10-7 [H+] =
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-Measures the concentration of hydrogen ions-the power/potential of the hydrogen ion-a way of expressing H+ in a given solution
The pH Scale
pH = - log [H+]
The higher the H+ concentration, the lower the pH
pOH = - log [OH-]
The higher the OH- concentration, the lower the pOH
pOH + pH = 14
18-17
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Figure 18.5
The pH values of some familiar
aqueous solutions
pH = -log [H3O+]
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Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.
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SAMPLE PROBLEM 18.3: Calculating [H3O+], pH, [OH-], and pOH
PROBLEM: Given [H3O+] concentrations: 2.0M, 0.30M, and 0.0063M, Calculate pH, [OH-], and pOH of the three solutions at 250C.
SOLUTION:
PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-] and then convert to pH and pOH.
For [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48
For [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80
For [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH
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Strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-(aq)
STRONG ACIDS and BASES
Strong acids and bases dissociate completely into ions in water.
100% ionization
18-21
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Strong Acids Strong Bases
HCl LiOH
HBr NaOH
HI KOH
HNO3 Ca(OH)2
H2 SO4 Sr(OH)2
HClO4 Ba(OH)2
All other acids and bases not listed here are considered weak.
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For STRONG ACIDS,
H+ or OH- are equal to the original molarity of the solution
HCl(g or l) + H2O(l) H3O+(aq) + Cl-(aq)
0.5 M 0.5 M0.5 M
Problem: Calculate the pH of 0.5 M HCl and 0.10 M NaOH solution:
pH = - log (0.5 M) = 0.30
NaOH(aq) Na+(aq) + OH-(aq)
0.10 M 0.10 M 0.10 M
pOH = - log (0.10 M) = 1.00pH = 14 – 1.00 = 13.00
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SAMPLE PROBLEM 18.5:
Calculate the pH of a solution prepared by dissolving 5.00 g of KOH in enough water to get 500.0 mL of solution
MW of KOH = 56.108 g/mol
Answer: 0.178 M of [OH-] pOH = 0.74902 pH = 13.25
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WEAK ACIDS and WEAK BASES
Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq)
Weak acids and bases dissociate very slightly into ions in water.
- About 1% ionization, the equilibrium lies far to the left
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The Acid-Dissociation Constant, Ka
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Kc = [H3O+][A-]
[HA]
Ka =[H3O+][A-]
[HA]
stronger acid higher [H3O+]
larger Ka
smaller Ka lower [H3O+]
weaker acid
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SAMPLE PROBLEM :
Write the equilibrium constant expression for the following weak acid:
a. NH4 +
NH4 + (aq) + H2O(l) H3O+(aq) + NH3
Ka =[H3O+][NH3]
[NH4 + ]
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Table 18.3 The Relationship Between Ka and pKa
Acid Name (Formula) Ka at 250C pKa
Hydrogen sulfate ion (HSO4-) 1.02x10-2
Nitrous acid (HNO2)
Acetic acid (CH3COOH)
Hypobromous acid (HBrO)
Phenol (C6H5OH)
7.1x10-4
1.8x10-5
2.3x10-9
1.0x10-10
1.991
3.15
4.74
8.64
10.00
pKa = - log Ka
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The Base-Dissociation Constant, Kb
A- (aq) + H2O(l) OH-(aq) + HA (aq)
Kc = [OH-][HA]
[A-] stronger base higher [OH-]
larger Kb
Kb = [OH-][HA]
[A-]
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SAMPLE PROBLEM :
Write the equilibrium constant expression for the following weak base:
a. C5H5N
C5H5N (aq) + H2O(l) OH-(aq) + C5H5NH+
Kb =[C5H5NH+ ][OH-]
[C5H5N]
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The Relationship Between Ka and Kb
Ka x Kb = Kw
Example: Calculate Kb if Ka = 1.5 x 10-5
Kb =1.0 x 10-14
1.5 x 10-5
= 6.7 x 10-10
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Lewis Definition of Acids and Bases
F
B
F F
H
N
H H
+
F
B
F F
H
N
H H
acid base adduct
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
M2+
H2O(l)
M(H2O)42+(aq)
adduct
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SAMPLE PROBLEM 18.13: Identifying Lewis Acids and Bases
PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions:
(a) H+ + OH- H2O
(b) Cl- + BCl3 BCl4-
(c) K+ + 6H2O K(H2O)6+
SOLUTION:
PLAN: Look for electron pair acceptors (acids) and donors (bases).
(a) H+ + OH- H2Oacceptor
donor
(b) Cl- + BCl3 BCl4-
donor
acceptor
(c) K+ + 6H2O K(H2O)6+
acceptor
donor